TITLE: Find a point on a parabola that's closest to another point.
QUESTION [7 upvotes]: Find the point on the parabola $3x^2+4x-8$ that is closest to the point $(-2,-3)$.
My plan for this problem was to use the distance formula and then that the derivative to get my answer. I'm having a little trouble along the way.
$$ d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.$$

REPLY [3 votes]: Suppose the closest point is at $p=(x_0,y_0)$, and set $q=(-2,-3)$. Then the tangent to the parabola at $p$ is perpendicular to $\ell$, the line through $p,q$.
Since $y'=6x+4$, the slope of the tangent is $6x_0+4$, so the slope of $\ell$ is $-\frac{1}{6x_0+4}$. Since it passes through $q$, we see the equation for $\ell$ is
$$y+3 = -\frac{x+2}{6x_0+4}$$
Combining with $y=3x^2+4x-8$ shows that $x_0$ is a root of
$$
(6x+4)(3x^2+4x-5)+(x+2)=18x^3+36x^2-13x-18=0
$$
and we end up numerically computing the roots from here.